Lecture 18

Inverse Probability Weighting

None of these models fits the data well.

\[ \begin{aligned} \mathcal{M}_1 &= \qty{ m(w,x) = a + bx \ : \ \text{ any numbers }\ a,b } && \text{ one line } \\ \mathcal{M}_2 &= \qty{ m(w,x) = a(w) : \ \text{ any function }\ a(w) } && \text{ horizontal lines } \\ \mathcal{M}_3 &= \qty{ m(w,x) = a(w) + bx \ : \ \text{ any function }\ a(w) \ \text{ and number }\ b } && \text{ parallel lines } \\ \mathcal{M}_4 &= \qty{ m(w,x) = a(w) + b(w)x \ : \ \text{ any functions }\ a(w),\ b(w) } && \text{ lines } \\ \mathcal{M}_5 &= \qty{ m(w,x) = a(w) + b \operatorname*{1_{\le 65}}(x) \ : \ \text{ any function } a(w) \ \text{ and number }\ b } && \text{ additive piecewise-constant } \\ \mathcal{M}_6 &= \qty{ m(w,x) = a(w) + b(w) \operatorname*{1_{\le 65}}(x) \ : \ \text{ any functions } a(w),\ b(w) } && \text{ piecewise-constant } \end{aligned} \]

But one does give us an unbiased estimate of the ATT.

Why?

Let’s think about what we’re estimating: the ATT.
This is a weighted average of column means.

\[ \begin{aligned} \text{ATT} &= \frac{1}{m_1}\sum_{j:w_j=1} \qty{ \mu(1,x_j) - \mu(0,x_j)} \qfor \mu(w,x) = \frac{\sum_{j: w_j,x_j=w,x } y_j}{\sum_{j: w_j,x_j=w} 1} \qand m_w = \sum_{j: w_j=w} 1 \\ &= \sum_{x} \textcolor[RGB]{0,191,196}{p(1,x)} \qty{ \textcolor[RGB]{0,191,196}{\mu(1,x)} - \textcolor[RGB]{248,118,109}{\mu(0,x)}} \qfor p(w,x) = \frac{\sum_{j: w_j=w, x_j=x} 1}{\sum_{j: w_j=w} 1 } \\ &= \sum_{w,x} \alpha(w,x) \mu(w,x) \qfor \alpha(w,x) = \begin{cases}+p(1,x) & \text{ if } w=1 \\ -p(1,x) & \text{ if } w=0 \end{cases} \end{aligned} \]

What you may notice is that, because these proportions \(p(w,x)\) are piecewise-constant, they’re in the model.

A Claim

You’ll get an unbiased estimate of the summary \(\sum_{w,x}\alpha(w,x) \mu(w,x)\)
if the regression model \(\mathcal{M}\) contains at least one of …
- the subpopulation mean function \(\mu(w,x)\)
- the weights \(\alpha(w,x)\)

Proof Ideas

Remember that our sampling distribution’s mean is our summary with the population least squares fit \(\tilde\mu\) plugged in for \(\mu\). \[ \mathop{\mathrm{E}}\qty[\widehat{\text{ATT}}] = \widetilde{\text{ATT}} := \textcolor[RGB]{0,191,196}{p(1,x)} \qty{ \textcolor[RGB]{0,191,196}{\tilde \mu(1,x)} - \textcolor[RGB]{248,118,109}{\tilde \mu(0,x)}} \]
Think about what we know about population least squares residuals.

\[ \begin{aligned} \frac{1}{n}\sum_{i=1}^n \qty{Y_i - \hat\mu(W_i,X_i)} m(W_i,X_i) &= 0 \qqtext{ for every } m \in \mathcal{M}&& \text{\textcolor{gray}{(sample least squares residuals)}} \\ \mathop{\mathrm{E}}\qty[ \qty{Y_i - \tilde\mu(W_i,X_i)} m(W_i,X_i)] &= 0 \qqtext{ for every } m \in \mathcal{M}&& \text{\textcolor{gray}{(population least squares residuals)}} \\ \end{aligned} \]

Weighted Least Squares

This sort of feels like an ‘a stopped clock is right twice a day’ thing.
We chose a ‘bad model’ for \(\mu\) but it worked because it was a good model for \(\alpha\).
But we can make this more systematic by using weighted least squares.

\[ \small{ \begin{aligned} \hat \mu = \mathop{\mathrm{argmin}}_{m \in \mathcal{M}} \frac{1}{n}\sum_{i=1}^n \textcolor[RGB]{0,0,255}{\gamma(W_i,X_i)} \qty{Y_i - m(W_i,X_i)}^2 \ \text{ satisfies } \ &\frac{1}{n}\sum_{i=1}^n \textcolor[RGB]{0,0,255}{\gamma(W_i,X_i)} \qty{Y_i - \hat\mu(W_i,X_i)} m(W_i,X_i) = 0 && \ \text{ for every } \ m \in \mathcal{M}\\ \tilde\mu = \mathop{\mathrm{argmin}}_{m \in \mathcal{M}} \mathop{\mathrm{E}}\qty[ \textcolor[RGB]{0,0,255}{\gamma(W_i,X_i)} \qty{Y_i - m(W_i,X_i)}^2 ] \ \text{ satisfies } \ &\mathop{\mathrm{E}}\qty[ \textcolor[RGB]{0,0,255}{\gamma(W_i,X_i)} \qty{Y_i - \tilde\mu(W_i,X_i)} m(W_i,X_i)] = 0 && \ \text{ for every } \ m \in \mathcal{M} \end{aligned} } \]

Our weights should reflect our summary.
e.g. if we take \(\gamma(w,x) = \abs{\alpha(w,x)}\), we’re good if our model contains \(m(w,x) = \text{sign}(\alpha(w,x))\).
For treatment effects, that means — if we choose the right weights — we just need to have indicators for treatment.

Inverse Probability Weighting

This just sort of worked out for us ‘by chance’ with our block-randomized experiment.
But let’s think about how to get the right weights in general. What are we doing?
We’re averaging the the red and green curves over the distribution of the green points.

\[ \]

\[ \small{\begin{aligned} \text{ATT} &= \sum_{x} \textcolor[RGB]{0,191,196}{p(1,x)} \qty{ \textcolor[RGB]{0,191,196}{\mu(1,x)} - \textcolor[RGB]{248,118,109}{\mu(0,x) }} \\ &= \underset{\text{\textcolor[RGB]{0,191,196}{matched} \textcolor{gray}{term}}}{\sum_{x} \textcolor[RGB]{0,191,196}{p(1,x) \tilde\mu(1,x)}} - \underset{\text{\textcolor[RGB]{0,191,196}{misma}\textcolor[RGB]{248,118,109}{tched} \textcolor{gray}{term}}}{\sum_{x} \textcolor[RGB]{0,191,196}{p(1,x)} \textcolor[RGB]{248,118,109}{\tilde\mu(0,x)}} \\ \end{aligned}} \]

\[ \small{\begin{aligned} \widetilde{\text{ATT}} &= \sum_{x} \textcolor[RGB]{0,191,196}{p(1,x)} \qty{ \textcolor[RGB]{0,191,196}{\tilde\mu(1,x)} - \textcolor[RGB]{248,118,109}{\tilde\mu(0,x) }} \\ &= \underset{\text{\textcolor[RGB]{0,191,196}{matched} \textcolor{gray}{term}}}{\sum_{x} \textcolor[RGB]{0,191,196}{p(1,x) \tilde\mu(1,x)}} - \underset{\text{\textcolor[RGB]{0,191,196}{misma}\textcolor[RGB]{248,118,109}{tched} \textcolor{gray}{term}}}{\sum_{x} \textcolor[RGB]{0,191,196}{p(1,x)} \textcolor[RGB]{248,118,109}{\tilde\mu(0,x)}} \\ \end{aligned}} \]

\[ \small{\begin{aligned} \text{bias} &= \widetilde{\text{ATT}} - \text{ATT} \\ &= \underset{\text{\textcolor[RGB]{0,191,196}{matched} \textcolor{gray}{term}}}{\textcolor{lightgray}{\sum_x \quad ? \quad}} - \underset{\text{\textcolor[RGB]{0,191,196}{misma}\textcolor[RGB]{248,118,109}{tched} \textcolor{gray}{term}}}{\textcolor{lightgray}{\sum_x \quad ? \quad }} \end{aligned}} \]

Does the Matched Term Cause Any Trouble?

\[ \text{bias} = \sum_{x} \textcolor[RGB]{0,191,196}{p(1,x) \qty{\tilde\mu(1,x) - \mu(1,x)}} \textcolor{lightgray}{ - \sum_{x} p(1,x) \qty{\tilde\mu(0,x)-\mu(0,x)}} \]

If our model has treatment-group indicators, no.

\[ \begin{aligned} 0 &= \mathop{\mathrm{E}}\qty[ \qty{Y_i - \tilde\mu(W_i,X_i)} 1(W_i=1) ] \\ &=\mathop{\mathrm{E}}\qty[ \qty{\mu(W_i,X_i) - \tilde\mu(W_i,X_i)} 1(W_i=1) ] \\ &=\mathop{\mathrm{E}}\qty[ \qty{\mu(1,X_i) - \tilde\mu(1,X_i)} 1(W_i=1) ] \\ &=\mathop{\mathrm{E}}\qty[ \qty{\mu(1,X_i) - \tilde\mu(1,X_i)} \mid W_i=1 ] P(W_i=1) \\ &=\sum_{x} \textcolor[RGB]{0,191,196}{p(1,x) \qty{\mu(1,x) - \hat\mu(1,x)}} \times P(W_i=1) \end{aligned} \]

Intuitively, this is because we’re ‘fitting the data in the right place’
We’re averaging over the same distribution of points we’ve fit our curve to.
So, even if our model is ‘bad’ most places, its average is still right.

What About the Mismatched Term?

\[ \text{bias} = \textcolor{lightgray}{\sum_{x} \tilde p(1,x) \qty{ \tilde\mu(1,x) - \mu(1,x)} -} \sum_{x} \textcolor[RGB]{0,191,196}{\tilde p(1,x)} \textcolor[RGB]{248,118,109}{\qty{\tilde\mu(0,x) - \mu(0,x)}} \]

Even when our model has treatment-group indicators, it’s still trouble. Let’s try what we just did.

\[ \small{\begin{aligned} 0 &= \mathop{\mathrm{E}}\qty[ \qty{Y_i - \tilde\mu(W_i,X_i)} 1(W_i=0) ] \\ &=\mathop{\mathrm{E}}\qty[ \qty{\mu(W_i,X_i) - \tilde\mu(W_i,X_i)} 1(W_i=0) ] \\ &=\mathop{\mathrm{E}}\qty[ \qty{\mu(0,X_i) - \tilde\mu(0,X_i)} 1(W_i=1) ] \\ &=\mathop{\mathrm{E}}\qty[ \qty{\mu(0,X_i) - \tilde\mu(0,X_i)} \mid W_i=0 ] P(W_i=0) \\ &= \sum_{x} \textcolor[RGB]{248,118,109}{p(0,x) \qty{\mu(0,x) - \tilde\mu(0,x)}} \times P(W_i=0) \end{aligned}} \]

This is not the right average. It’s the one that occurs in the ‘unadjusted gap’, not the ATT.
We need to adjust for the fact that we’re averaging over a different distribution of points.

The Second: A Weighted Version

Recall what we know about weighted least squares and its residuals. \[ \tilde \mu = \mathop{\mathrm{argmin}}_{m \in \mathcal{M}} \mathop{\mathrm{E}}\qty[ \gamma(W_i,X_i) \qty{Y_i - m(W_i,X_i)}^2 ] \ \text{ satisfies } \ \mathop{\mathrm{E}}\qty[ \gamma(W_i,X_i) \qty{Y_i - \tilde\mu(W_i,X_i)} m(W_i,X_i)] = 0 \ \text{ for every } \ m \in \mathcal{M} \]
Let’s try ‘upweighting’ the red points on the right side by the ratio of the number of green points to red ones.

\[ \small{\begin{aligned} 0 &= \mathop{\mathrm{E}}\qty[ \gamma(X_i) \qty{Y_i - \tilde\mu(W_i,X_i)} 1(W_i=0) ] \qfor \gamma(x)= \frac{\textcolor[RGB]{0,191,196}{p(1,x)}}{\textcolor[RGB]{248,118,109}{p(0,x)}} \\ &=\mathop{\mathrm{E}}\qty[ \gamma(X_i)\qty{\mu(W_i,X_i) - \tilde\mu(W_i,X_i)} 1(W_i=0) ] \\ &=\mathop{\mathrm{E}}\qty[ \gamma(X_i)\qty{\mu(0,X_i) - \tilde\mu(0,X_i)} 1(W_i=1) ] \\ &=\mathop{\mathrm{E}}\qty[ \gamma(X_i)\qty{\mu(,X_i) - \tilde\mu(0,X_i)} \mid W_i=0 ] \textcolor{lightgray}{P(W_i=0)} \\ &= \sum_{x} \textcolor[RGB]{248,118,109}{p(0,x)} \gamma(0,x) \textcolor[RGB]{248,118,109}{\qty{\mu(0,x) - \tilde\mu(0,x)}} \textcolor{lightgray}{\times P(W_i=0)} \\ &= \sum_{x} \textcolor[RGB]{248,118,109}{p(0,x)} \frac{\textcolor[RGB]{0,191,196}{p(1,x)}}{\textcolor[RGB]{248,118,109}{p(0,x)}} \textcolor[RGB]{248,118,109}{\qty{\mu(0,x) - \tilde\mu(0,x)}} \textcolor{lightgray}{\times P(W_i=0)}\\ &= \sum_{x} \textcolor[RGB]{0,191,196}{p(1,x)} \textcolor[RGB]{248,118,109}{\qty{\mu(0,x) - \tilde\mu(0,x)}} \textcolor{lightgray}{\times P(W_i=0)} \end{aligned}} \]

This works. Intuitively, using these weights means that we’re ‘fitting the data in the right place’ again.

Summary

\[ \begin{aligned} \widehat{\text{ATT}} = \sum_{x} \textcolor[RGB]{0,191,196}{\hat p(1,x)} \qty{ \textcolor[RGB]{0,191,196}{\hat\mu(1,x)} - \textcolor[RGB]{248,118,109}{\hat\mu(0,x)}} \qfor \hat \mu = \mathop{\mathrm{argmin}}_{m \in \mathcal{M}} \sum_{i=1}^n \gamma(W_i,X_i) \qty{Y_i - m(W_i,X_i)}^2 \qfor \frac{\textcolor[RGB]{0,191,196}{p(1,x)}}{p(w,x)} \end{aligned} \]

This is the ‘inverse probability weighted least squares’ estimate of the ATT.
We give equal weight to the green points because that’s the distribution we’re averaging over.
But we weight the red points by the ratio of the number of green points to red ones so that …
- if we ‘count them by weight’ …
- then they look like they have that distribution too.

Summary

\[ \small{ \begin{aligned} \widehat{\text{ATT}} = \sum_{x} \textcolor[RGB]{0,191,196}{\hat p(1,x)} \qty{ \textcolor[RGB]{0,191,196}{\hat\mu(1,x)} - \textcolor[RGB]{248,118,109}{\hat\mu(0,x)}} \qfor \hat \mu = \mathop{\mathrm{argmin}}_{m \in \mathcal{M}} \sum_{i=1}^n \gamma(W_i,X_i) \qty{Y_i - m(W_i,X_i)}^2 \qfor \frac{\textcolor[RGB]{0,191,196}{p(1,x)}}{p(w,x)} \end{aligned} } \]

model	true	unweighted	weighted
A	0.24	0.16	0.25
B	0.24	0.00	0.00
C	0.24	0.24	0.25
D	0.24	-0.06	0.25
E	0.24	0.17	0.25
F	0.24	0.24	0.25

This is the ‘inverse probability weighted least squares’ estimate of the ATT.
We give equal weight to the green points because that’s the distribution we’re averaging over.
But we weight the red points by the ratio of the number of green points to red ones so they look like they have that distribution too.

This Works For Any Distribution of Assignments

model	true	unweighted	weighted
A	0.23	0.16	0.25
B	0.23	0.00	0.00
C	0.23	0.16	0.22
D	0.23	-0.03	0.25
E	0.23	0.10	0.25
F	0.23	0.23	0.24